RACA Journal September 2022 | Seite 45

Support used to melt the ice . Theoretically the water should stay at 0 ° C , however because the transfer of heat to the ice is slow the water will increase in temperature as well . The heat added at this stage is mostly “ latent heat ” because a change in phase is taking place .

Once all the ice has melted , the water will increase in temperature ; the heat added now is “ sensible heat ”. It can be sensed , or felt . When the water reaches boiling point ( which depends on the altitude ) the temperature will remain constant and the water will start to evaporate . Because now there is no increase in temperature but a change in phase the heat added is “ latent heat ”.

Example 5 Question : A pot of water containing 1 litre of water at 0 ° C and 1kg of ice , also at 0 ° C , is put on a stove . We are at sea level altitude . Calculate the amount of heat required to produce steam at 110 ° C . The boiling point of the water at sea level is 100 ° C . The heat of fusion of ice is 333,7 kJ / kg . The specific heat of water is 4,2 kJ / kg and the latent heat of vaporisation of water at 100 ° C is 2 257 kJ / kg . The specific heat of water vapour is 1,89 kJ / kg .

Answer : It is obvious that there are changes of phase and changes in temperature involved . The sequence of events is as follows : 1 . The ice melts . This is latent heat . 2 . The water is heated from 0 ° C to 100 ° C . This is sensible heat . 3 . The water evaporates at 100 ° C . This is latent heat . 4 . The water vapour is heated from 100 to 110 ° C . This is again sensible heat .

Calculations : To melt 1kg of ice requires 1kg X 333,7kJ / kg = 333,7kJ

To evaporate the water requires 2kg X 2257kJ = = 4 514,0kJ To heat the water vapour from 100 to 110 ° C requires 2 X 1,89 X ( 110 – 100 K ) = 37,8kJ .

----------------- Total amount of heat required = 5 725,5kJ

Example 6 Question : A humidifier must add 5kg of steam per hour to an air stream . If the water being supplied to the humidifier is 20 ° C and the boiling point of water at the site is 95 ° C . Calculate the size of element required to produce this steam if the latent heat of vaporisation at this temperature is 2270 kJ / kg .

Answer : To produce 5kg of steam 5kg of water is required . This water first has to be heated from 20 ° C to 95 ° C and then evaporated . The amounts of heat required are as follows : To heat the water : 5 X 4,2 X ( 95 – 20 ) = 1575kJ To evaporate the water requires 5 X 2270 = 11 350kJ Total amount of heat required = 12 925 kJ

This amount of heat must be added in 1 hour ; thus the amount to be added per second is : 12 925 ÷ 3600 = 3,59 kJ / s . This is equivalent to 3,59 kW .

Note : the size of the electrical element is the amount of heat it can add per second .

Example 7 Question : Calculate the refrigerating capacity of a plant which produces 2000kg of ice in an 8-hour shift if the water supply is at 20 ° C and the ice is required at -10 ° C .

Answer : For 2000 kg of ice , 2000 kg of water is required . This water must be cooled from 20 ° C to 0 ° C . The heat to be removed for this = 2000 X 4,2 X ( 20 – 0 ) = 168 000 kJ Next this water must be changed to ice . The heat to be removed for this = 2000 X 333,7 = 667 400kJ Next the ice must be cooled from 0 ° C to -10 ° C . The heat to be removed for this = 2000 X 2,1 X ( 0 - -10 ) = 42 000kJ

------------- Total amount of heat to be removed in 8 hours = 877 400kJ

The capacity of the plant is the amount of heat it must be able to remove per second .

This is calculated by dividing the total amount of heat to be removed by the number of seconds in the time allowed .

Thus the capacity of the plant is : 877 400 ÷ ( 8 x 3600 ) = 877 400 ÷ 28 800 = 30,47 kJ / s = 30,47 kW

Thank you for the question Stefan , I hope that this sheds some light on the fundamentals of refrigeration .

Thank you for all your questions . Send your problems ( and sometimes your creative solutions ) to acra @ netactive . co . za with ‘ Solutions Page ’ in the subject line . You may include pictures . RACA